Advertisements
Advertisements
प्रश्न
How many quadratic equations can be formed using numbers from 0, 2, 4, 5 as coefficient if a coefficient can be repeated in an equation.
Advertisements
उत्तर
Let the quadratic equation be
ax2 + bx + c = 0, a ≠ 0
| Coefficient | Values | Numbers of ways |
| a | 2, 4, 5 | 3 |
| b | 0, 2, 4, 5 | 4 |
| c | 0, 2, 4, 5 | 4 |
Required number = 3 × 4 × 4 = 48
APPEARS IN
संबंधित प्रश्न
A teacher wants to select the class monitor in a class of 30 boys and 20 girls, in how many ways can the monitor be selected if the monitor must be a boy? What is the answer if the monitor must be a girl?
How many three-digit numbers can be formed using the digits 2, 3, 4, 5, 6 if digits can be repeated?
How many five-digit numbers formed using the digit 0, 1, 2, 3, 4, 5 are divisible by 3 if digits are not repeated?
Compute: `(12/6)!`
Compute: `(6! - 4!)/(4!)`
Compute: `(8!)/(6! - 4!)`
Compute: `(8!)/((6 - 4)!)`
Write in terms of factorial:
3 × 6 × 9 × 12 × 15
Write in terms of factorial:
6 × 7 × 8 × 9
Write in terms of factorial:
5 × 10 × 15 × 20 × 25
Find n, if `"n"/(8!) = 3/(6!) + 1/(4!)`
Find n if: `("n"!)/(3!("n" - 3)!) : ("n"!)/(5!("n" - 5)!)` = 5:3
Find n if: `("n"!)/(3!("n" - 5)!) : ("n"!)/(5!("n" - 7)!)` = 10:3
Find n, if: `((2"n")!)/(7!(2"n" - 7)!) : ("n"!)/(4!("n" - 4)!)` = 24:1
Find the value of: `(5(26!) + (27!))/(4(27!) - 8(26!)`
A hall has 12 lamps and every lamp can be switched on independently. Find the number of ways of illuminating the hall.
